Difference between revisions of "1953 AHSME Problems/Problem 8"

(Created page with "==Problem 8== The value of <math>x</math> at the intersection of <math>y=\frac{8}{x^2+4}</math> and <math>x+y=2</math> is: <math>\textbf{(A)}\ -2+\sqrt{5} \qquad \textbf{(B...")
 
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== Solution ==
 
== Solution ==
 
<math>x+y=2\implies y=2-x</math>. Then <math>2-x=\frac{8}{x^2+4}\implies (2-x)(x^2+4)=8</math>. We now notice that <math>x=0\implies (2)(4)=8</math>, so <math>\fbox{C}</math>
 
<math>x+y=2\implies y=2-x</math>. Then <math>2-x=\frac{8}{x^2+4}\implies (2-x)(x^2+4)=8</math>. We now notice that <math>x=0\implies (2)(4)=8</math>, so <math>\fbox{C}</math>
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==See Also==
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{{AHSME 50p box|year=1953|num-b=7|num-a=9}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 20:40, 1 April 2017

Problem 8

The value of $x$ at the intersection of $y=\frac{8}{x^2+4}$ and $x+y=2$ is:

$\textbf{(A)}\ -2+\sqrt{5} \qquad \textbf{(B)}\ -2-\sqrt{5} \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ \text{none of these}$

Solution

$x+y=2\implies y=2-x$. Then $2-x=\frac{8}{x^2+4}\implies (2-x)(x^2+4)=8$. We now notice that $x=0\implies (2)(4)=8$, so $\fbox{C}$

See Also

1953 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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